cal_days_in

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sasha at alronix dot com
05-Apr-2006 09:46


function days_in_month($a_month, $a_year) {

   return date('t', strtotime($a_year . '-' . $a_month . '-01'));

}

NiGhMa
23-May-2005 04:00


Stay simple ;)



$ndays = date("t"); //return the number of days for this month and this year



$ndays = date("t", mktime(0, 0, 0, 12, 1, 2004)); // You can also specify a date 



c u

Pasci - webmaster at cyber-style dot net
07-Jan-2005 06:21


Here's the one line solution without loops or other slow stuff ;)



<?php



$daynum = date("j",mktime(0,0,0,$month+1,0,$year));



?>

richard at N O S P A M dot computoz dot co dot uk
24-Aug-2004 08:55


re: rshowalter at mtsinai dot on dot ca 's daysinmonth.



This is faster (no loop, no testing irrelevant values):



    function daysinmonth($month, $year)

    {

        if(checkdate($month, 31, $year)) return 31;

        if(checkdate($month, 30, $year)) return 30;

        if(checkdate($month, 29, $year)) return 29;

        if(checkdate($month, 28, $year)) return 28;

        return 0; // error

    }



and won't inf-loop on you, ever.

rshowalter at mtsinai dot on dot ca
28-Apr-2004 08:43


How about this one:





<?php


function daysinmonth($month, $year)


{


    $days=31;


    while(!checkdate($month, $days, $year)) $days--;


    return $days;


}


?>





Not exactly a one-line solution, but easy to follow, not limited to 1970-2038 (but don't use it after the year 32767) and uses a built-in function that's always available.





Be sure that $month and $year are valid, or the loop will run forever. Possibly add:





&& $days>0





to the while condition. This would return 0 as your error code.

paul (at) bughunt3r.com
18-Apr-2004 11:31


Here is a nifty function to return the date for easter for ranges beyond 2037 or before 1970.



<?php

    function str_easter_date ($year) {

        $days = easter_days ($year);

        $days_in_march = cal_days_in_month (CAL_GREGORIAN, 3, $year);

        if (($days + 21) < $days_in_march) {

            $month = "March";

            $date = $days + 21;

        } else {

            $month = "April";

            $date = ($days + 21) - $days_in_march;

        }

        

        return "$month $date, $year";

    }

?>

jeffbeall at comcast dot net
15-Apr-2004 01:04


This will work great in future dates but will give the wrong answer for dates before 1550 (approx) when leap year was introduced and the calendar lost a year or two.

Sorry now to be more specific it has been a while sine I had to account for those later dates and had to take that into account but just a heads up for others to watch out.

dbindel at austin dot rr dot com
02-Jan-2004 01:06


Here's a one-line function I just wrote to find the numbers of days in a month that doesn't depend on any other functions.





The reason I made this is because I just found out I forgot to compile PHP with support for calendars, and a class I'm writing for my website's open source section was broken. So rather than recompiling PHP (which I will get around to tomorrow I guess), I just wrote this function which should work just as well, and will always work without the requirement of PHP's calendar extension or any other PHP functions for that matter.





I learned the days of the month using the old knuckle & inbetween knuckle method, so that should explain the mod 7 part. :)





<?php


/*


 * days_in_month($month, $year)


 * Returns the number of days in a given month and year, taking into account leap years.


 *


 * $month: numeric month (integers 1-12)


 * $year: numeric year (any integer)


 *


 * Prec: $month is an integer between 1 and 12, inclusive, and $year is an integer.


 * Post: none


 */


// corrected by ben at sparkyb dot net


function days_in_month($month, $year)


{


// calculate number of days in a month


return $month == 2 ? ($year % 4 ? 28 : ($year % 100 ? 29 : ($year % 400 ? 28 : 29))) : (($month - 1) % 7 % 2 ? 30 : 31);


}


?>





Enjoy,


David Bindel

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